Answer:
The angle formed between the line of pull of a muscle and the longitudinal axis of the bone in which the muscle is acting. The line of pull is usually indicated by the joint angle. It affects the strength of muscle action; at only certain angles of pull can a muscle exert maximal tension.
Explanation:
hope this helps sorry if it doesn't have a good rest of your night :) ❤
If the magnitude of vector A⃗ is less than the magnitude of vectorB⃗ , then the x component of A⃗ is less than the x component ofB⃗ . If the magnitude of vector is less than the magnitude of vector, then the component of is less than the component of.
a. True
b. False
pers
2. (a) Calculate the virtual depth of a black dot at the
bottom of a cubic block made of transparent glass with each
side 4 cm, while the refractive index of glass is 1.6.
Answer:
2.5 cm
Explanation:
Using the relation :
Refractive index = Real Depth / Apparent depth
Refractive index = 1.6
Real depth = 4cm
Virtual depth = apparent depth = x
1.6 = 4cm / x
1.6x = 4
x = 4 / 1.6
x = 2.5
Hence, virtual depth = 2.5cm
Please help! THIS IS A EASY ONE, HOPEFULLY.
Answer:
megaliter > kiloliter > liter >centiliter >mililiter > deciliter > nanoliter
Explanation:
plz mark brainlest
.................,,,,,,,,,,,
Answer:
B
Explanation:
Motion is movement, the teacher's movement is motion
The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1300 N/m that he will compress with a force of 6500 N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 50 N during the 5.0 mm he moves in the barrel.
Required:
At what speed will he emerge from the end of the barrel, 2.5 mabove his initial rest position?
Answer:
22m/s
Explanation:
Mass, m=60 kg
Force constant, k=1300N/m
Restoring force, Fx=6500 N
Average friction force, f=50 N
Length of barrel, l=5m
y=2.5 m
Initial velocity, u=0
[tex]F_x=kx[/tex]
Substitute the values
[tex]6500=1300x[/tex]
[tex]x=\frac{6500}{1300}=5[/tex]m
Work done due to friction force
[tex]W_f=fscos\theta[/tex]
We have [tex]\theta=180^{\circ}[/tex]
Substitute the values
[tex]W_f=50\times 5cos180^{\circ}[/tex]
[tex]W_f=-250J[/tex]
Initial kinetic energy, Ki=0
Initial gravitational energy, [tex]U_{grav,1}=0[/tex]\
Initial elastic potential energy
[tex]U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)[/tex]
[tex]U_{el,1}=16250J[/tex]
Final elastic energy,[tex]U_{el,2}=0[/tex]
Final kinetic energy, [tex]K_f=\frac{1}{2}(60)v^2=30v^2[/tex]
Final gravitational energy, [tex]U_{grav,2}=mgh=60\times 9.8\times 2.5[/tex]
Final gravitational energy, [tex]U_{grav,2}=1470J[/tex]
Using work-energy theorem
[tex]K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}[/tex]
Substitute the values
[tex]0+0+16250-250=30v^2+1470+0[/tex]
[tex]16000-1470=30v^2[/tex]
[tex]14530=30v^2[/tex]
[tex]v^2=\frac{14530}{30}[/tex]
[tex]v=\sqrt{\frac{14530}{30}}[/tex]
[tex]v=22m/s[/tex]
toy car A drives with a steady force of 35N and covers 2000 m with fully charged battery. toy car B drives with a steady force of 80 N. how far would it be able to drive using the same fully charged battery as car A.
The distance travelled by toy car B using the same fully charged battery as car A is 875 m
How to determine the energy of car AForce (F) = 35 NDistance of car A (d) = 2000 mEnergy (E) = ?E = fd
E = 35 × 2000
E = 70000 J
How to determine the distance travelled by car BEnergy (E) = 70000 JForce (F) = 80 NDistance of car B =?E = fd
70000 = 80 × Distance of car B
Divide both sides by 80
Distance of car B = 70000 / 80
Distance of car B = 875 m
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The magnitude obtained when adding vector A (80 N at 20 deg) with vector B (40 N at
70 deg) is:
110.06 N
89.85 N
0 130.32 N
0 141.98 N
Answer:
110.06NExplanation:
The magnitude of the force is known as the resultant.
R = √Fx²+Fy²
Fx = 80cos 20 + 40cos70
Fx = 80(0.9397)+40(0.3420)
Fx = 75.176 + 13.68
Fx = 88.856N
Fy = 80sin 20 + 40sin70
Fy = 80(0.3420)+40(0.9397)
Fy = 27.36 + 37.588
Fy = 64.948N
R = √88.586²+64.948²
R = √7,847.48+4,218.24
R = √12,065.72
R = 109.5
R = 110N
Hence the magnitude of the forces is 110N
coefficient of static friction formula
If the mass of the object doubles then the acceleration is when the force is kept the same
Answer:
Halved
Explanation:
F=ma
Let case 1 (original) be:
[tex]F_{1}=m_{1} a_{1} \\[/tex]
Case 2 (new) be:
[tex]F_{2}=m_{2} a_{2}[/tex]
Mass is double:
[tex]m_{2}= 2m_{1}[/tex]
Force kept the same:
[tex]F_{1} =F_{2}[/tex]
Combine the equation and gives:
[tex]\frac{F_{1} }{F_{2}} =\frac{m_{1} a_{1} }{m_{2}a_{2} }\\\frac{F_{1} }{F_{1}} =\frac{m_{1} a_{1} }{2m_{1}a_{2} }\\1=\frac{a_{1} }{2a_{2} }\\a_{2}=\frac{1}{2} a_{1}[/tex]
Acceleration is halved
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is doubly ionized. Calculate the distance from the uranium atom at which an electron will be in equilibrium. Ignore the gravitational attraction between the particles. = nm An electron sits between a singly ionized uranium ion and a doubly ionized iron ion. The distance from the uranium ion to the electron is designated lowercase r, and the distance between the two ions is designated uppercase R. What is the magnitude of the force on the electron from the uranium ion? magnitude of the force: N
Answer:
Explanation:
Charge on uranium ion = charge of a single electron
= 1.6 x 10⁻¹⁹ C
charge on doubly ionised iron atom = charge of 2 electron
= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C
Let the required distance from uranium ion be d .
force on electron at distance d from uranium ion
= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²
force on electron at distance 61.10 x 10⁻⁹ - r from iron ion
= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²
For equilibrium ,
9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²
2 d² = (61.10 x 10⁻⁹ - r )²
1.414 r = 61.10 x 10⁻⁹ - r
2.414 r = 61.10 x 10⁻⁹
r = 25.31 nm .
(a) The distance from the uranium atom at which an electron will be in equilibrium is [tex]2.04 \times 10^{-8} \ m[/tex]
(b) The magnitude of the force on the electron from the uranium ion is [tex]3.46 \times 10^ 6 \ N[/tex]
The given parameters:
distance between the iron and the uranium, d = 61.1 nmcharge of uranium ion, q₁ = 1.6 x 10⁻¹⁹ Ccharge of doubly ionized atom, q₂ = 2q₁ = 3.2 x 10⁻¹⁹ CThe force on the electron due to uranium ion at distance r is calculated as follows;
[tex]F _1 = \frac{Kq_1^2}{r^2} \\\\F_1 = \frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{r^2} \\\\F_1 = \frac{2.3 \times 10^{-28}}{r^2}[/tex]
The force on the electron due to uranium ion at distance less than 61.10 nm.
R = 61.10 nm - r
[tex]F_2 = \frac{9\times 10^9 \times (3.2 \times 10^{-19})^2}{(61.1 \times 10^{-9} \ - \ r)^2} \\\\\F_2 = \frac{9.216 \times 10^{-28}}{(61.1 \times 10^{-9} \ - \ r)^2}[/tex]
At equilibrium, the force between the electron and ions will be equal.
[tex]\frac{9.216 \times 10^{-28}}{(61.1 \times 10^{-9} \ - \ r)^2}= \frac{2.3 \times 10^{-28}}{r^2}\\\\\frac{4}{(61.1 \times 10^{-9} \ - \ r)^2} = \frac{1}{r^2} \\\\4r^2 = (61.1 \times 10^{-9} \ - \ r)^2\\\\2^2r^2 = (61.1 \times 10^{-9} \ - \ r)^2\\\\2r = 61.1 \times 10^{-9} \ - \ r\\\\3r = 61.1 \times 10^{-9} \\\\r = \frac{61.1 \times 10^{-9}}{3} \\\\r = 2.04 \times 10^{-8} \ m[/tex]
The magnitude of the force on the electron from the uranium ion is calculated as follows;
[tex]F = \frac{kq_1^2}{r^2} \\\\F = \frac{9\times 10^9 \times 1.6\times 10^{-19}}{(2.04 \times 10^{-8})^2} \\\\F= 3.46 \times 10^6 \ N[/tex]
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Two charges, one +Q and the other −Q, are held a distance d apart. Consider only points on the line passing through both charges and clearly explain your answers to the following: [You can answer this problem without any calculations]. Do not consider any points at infinite distance from the charges. [5 points](a) Find the location of all points, if any, where the electric potential is zero.(b) Find the location of all points, if any, where the electric field is zero.
Answer:
a. d/2 mid-way between the charges.
b. d/2 mid-way between the charges.
Explanation:
(a) Find the location of all points, if any, where the electric potential is zero.
Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric potential due to the +Q charge is V = kQ/x and that due to the -Q charge is V' = -kQ/(d - x) where x is the point of zero electric potential.
The potential is zero when V + V' = 0, and this can only be midway between the charges. This is shown below
So, kQ/x + [-kQ/(d - x)] = 0
kQ/x - kQ/(d - x) = 0
kQ/x = kQ/(d - x)
1/x = 1/(d - x)
(d - x) = x
d = x + x
d = 2x
x = d/2 which is mid-way between the charges.
(b) Find the location of all points, if any, where the electric field is zero.
Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric field due to the +Q charge is E = kQ/x² and that due to the -Q charge is E' = -kQ/(d - x)² where x is the point of zero electric field.
The electric field is zero when E + E' = 0 and this can only be midway between the charges. This is shown below.
So, kQ/x² + [-kQ/(d - x)²] = 0
kQ/x² - kQ/(d - x)² = 0
kQ/x² = kQ/(d - x)²
1/x² = 1/(d - x)²
(d - x)² = x²
d - x = ± x
d = x ± x
d = x - x or x + x
d = 0 or 2x
d = 0 or d = 2x
Since d ≠ 0, d = 2x ⇒ x = d/2 which is midway between the charges.
How much force will a 5 kg rock hit the Earth with if it falls
for 1 second?
Answer:
f
Explanation:
f
Which graph best represents the greatest amount of work
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.
This question is incomplete, the missing diagram is uploaded along this Answer below.
Answer:
a) the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight is - 3.935 lb-ft
Explanation:
Given the data in the question;
(a) determine the work done on the cart by the spring
we calculate the work done on the cart by the spring as follows;
[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )
where k is spring constant ( 3 lb/in )
we substitute
[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )
[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )
[tex]W_{spring}[/tex] = 1/2 × 3( 39 )
[tex]W_{spring}[/tex] = 58.5 lb-in
we convert to pound force-foot
[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft
[tex]W_{spring}[/tex] = 4.875 lb-ft
Therefore, the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight
work done by its weight;
[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )
we substitute in of values from the image below;
[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )
[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13
[tex]W_{gravity}[/tex] = -47.1 lb-in
we convert to pound force-foot
[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft
[tex]W_{gravity}[/tex] = - 3.935 lb-ft
Therefore, the work done on the cart by its weight is - 3.935 lb-ft
a) the work done on the cart by the spring is 4.875 lb-ft.
b) the work done on the cart by its weight is - 3.935 lb-ft.
Calculation of the work done:a. The work done on the cart by the spring is
= 1/2 × 3( (-8)² - (5)² )
= 1/2 × 3( 64 - 25 )
= 1/2 × 3( 39 )
= 58.5 lb-in
Now we have to convert to pound force-foot
So,
= 58.5 × 0.0833333 lb-ft
= 4.875 lb-ft
b) Now
work done by its weight;
= -mgsin∅( x₂ - x₁ )
So,
= -14 × sin(15°)( 5 - (-8) )
= -14 × 0.2588 × 13
= -47.1 lb-in
Now we convert to pound force-foot
= -47.1 × 0.0833333 lb-ft
= - 3.935 lb-ft
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what was Thomas Edison first major invented?
Answer:
Thomas Edisons most famous invention was the phonograph
Thomas Edison announces his invention of the phonograph, a way to record and play back sound. Edison stumbled on one of his great inventions—the phonograph—while working on a way to record telephone communication at his laboratory in Menlo Park, New Jersey.
Explanation:
Hope I helped
What force is needed to give a 4800.0 kg truck an acceleration of 6.2 m/s2 over a level road?
Answer:
the force needed to give the truck the acceleration is 29,760 N.
Explanation:
Given;
mass of truck, m = 4800 kg
acceleration of the truck, a = 6.2 m/s²
The force needed to give the truck the acceleration is calculated as;
F = ma
F = 4800 x 6.2
F = 29,760 N
Therefore, the force needed to give the truck the acceleration is 29,760 N.
A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two balls after collision?
Answer:
We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.
Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.
Explanation:
An object is placed in material a at point P, as shown in the diagram. The light is refracted when it strikes the interface with material b. When viewed from material b, at which point will the image appear?
An iron nail becomes a permanent magnet if it is
if you stroke it an iron nail with a bar magnet the nail will become a permanent or long lasting magnet.
Hope it's perfect for you.
According to the article, what was the effect of elevation on the experimental group?
Answer:
Due to the effect of elevation on the experimental group the participants decided to try to help the research assistant who was having opening one of her files to finish the study. Schnall, Roper, and Fessler were able to conclude that happiness associated with a feeling of elevation can lead to more altruism or helping behaviors.
Explanation:
100 percent on edge
The effect of elevation on the experimental group was they show more urge of being altruistic and feeling happiness and satisfaction associated with elevation.
Altruism:
It is a practice in which a person help others without any selfishness, the person just want to help.
Witnessing someone's altruistic behavior, make other to feel good and a urge of being Altruistic, this is known as elevation.
After watching elevating Oprah video, the test group help the research assistant who was having trouble in opening file.
Therefore, the effect of elevation on the experimental group was they show more urge of being altruistic and feeling happiness and satisfaction associated with elevation.
To know more about Altruism:
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An ideal heat engine operates between 778 K and 475 K. 267 J of waste heat is exhausted. What is the input heat?
Answer:
Explanation:
Suppose that the turbines of a coal-fired plant are driven by hot gases at a temperature of 886 K. the temperature of the exhaust area is only 305 K, the efficiency of this heat engine
HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!
A car with a mass of 1,200 kg accelerates at a rate of 3.0 m/s^2 forward. What is the force acting on the car?
Answer:
the force acting on the car is 3600 N
Explanation:
The computation of the force acting on the car is shown below:
As we know that
Force = mass × acceleration
= 1200 kg × 3.0 ms/^2
= 3600 N
hence, the force acting on the car is 3600 N
Balanced forces acting on an object keeps it at ____or moving at _____in straight line.
Fill in the ___ spaces
Answer:
Please see below as the answer is self explanatory.
Explanation:
According to Newton's 2nd law, a net force acting on an object of mass m, causes the object to be accelerated.If the forces acting on the object are balanced, which means that the net force on the object is zero, just applying the same law, we find that the object is not accelerated.According to Newton's First law, an object that is not accelerated is at rest, or moves along a straight line at constant speed.So, if there are balanced forces acting on the object, if the object is at rest, will keep at rest, and if it is moving, it will keep moving at constant speed along a straight line.What body system,
respiratory system
Circulatory system
Digestive system
Nervous system
Affects ulcer disease and heartburn.
Please do this quick it is due please. I will make brainlest and give out extra points
Affects ulcer disease and heartburn.
Please do this quick it is due please. I will make brainlest and give out extra points!
Answer:
Digestive system
Explanation:
ulcer affect anywhere in the digestive system
Digestive system.
Since the acids in your food break down with the chemicals in your stomach, it can give you heartburn and also, ulcer disease happens in your stomach so the only correct answer would be Digestive System. I would like to say that the person with the profile name BigPapa who commented on my answer deserves a lot of credit, and thanks if you see this.
-R3TR0 Z3R0
You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 4, 4, -3 > m/s, and the velocity of the other lump was < -2, 0, -7 > m/s.
What was the the total momentum of the lumps just before the impact?
p(total) = ____kg·m/s.
What is the momentum of the stuck-together lump just after the collision?
p = ____kg·m/s.
What is the velocity of the stuck-together lump just after the collision?
v_f = ____m/s.
Answer:
a) p(total) = <0.05, 0.1, 0.1 > kg m/s
b) p = <0.05, 0.1, 0.1 > kg.m/s
c) v_f = < 1, 2, 2 > m/s
Explanation:
a.)
Mass of each lump = 25 g = 0.025 kg
Velocity of lump 1 = < -2, 0, -7 > m/s
Momentum of lump 1 = Mass×Velocity
= 0.025×< -2, 0, -7 >
= < -0.05, 0, 0.175> kg m/s
Velocity of lump 2 = < 4, 4, -3 > m/s
Momentum of lump 2 = Mass×Velocity
= 0.025×< 4, 4, -3 >
= < 0.1, 0.1, -0.075> kg m/s
Total momentum before impact = < -0.05, 0, 0.175 > + < 0.1, 0.1, -0.075>
= < 0.05, 0.1, 0.1 > kg m/s
⇒p(total) = <0.05, 0.1, 0.1 > kg m/s
b)
As we know that,
By the law of conservation of linear momentum,
The total momentum will be the same before and after the collision.
⇒Momentum of the stuck together after the collision = Total momentum of the lumps just before impact.
⇒ p = <0.05, 0.1, 0.1 > kg m/s
c)
Let the final velocity = v_f
Total mass = 0.025 + 0.025 = 0.05 kg
As
Momentum = mass ×velocity
⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f
⇒ v_f = <0.05, 0.1, 0.1 > / 0.05
= < 1, 2, 2 > m/s
⇒v_f = < 1, 2, 2 > m/s
Name the nutrients required for the
body.
Answer:
1- water
2- fat
3- carbohydrates
4- vitamins
5- minerals
What did people assume Katherine was when she entered the room?
Answer: custodian,
Explanation: they never saw any colored women in the division before
A ball is thrown vertically downward from the top of a 37.4-m-tall building. The ball passes the top of a window that is 15.4 m above the ground 2.00 s after being thrown. What is the speed of the ball as it passes the top of the window?
Answer:
v= 20.8 m/s
Explanation:
Assuming no other forces acting on the ball, from the instant that is thrown vertically downward, it's only accelerated by gravity, in this same direction, with a constant value of -9.8 m/s2 (assuming the ground level as the zero reference level and the upward direction as positive).In order to find the final speed 2.00 s after being thrown, we can apply the definition of acceleration, rearranging terms, as follows:[tex]v_{f} = v_{o} + a*t = v_{o} + g*t (1)[/tex]
We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:[tex]\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)[/tex]
where Δy = yf - y₀ = 15.4 m - 37.4 m = -22 m (3)Replacing by the values of Δy, a and t, we can solve for v₀ as follows:[tex]v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)[/tex]
Replacing (4) , and the values of g and t in (1) we can find the value that we are looking for, vf:[tex]v_{f} = v_{o} + g*t = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)[/tex]
Therefore, the speed of the ball (the magnitude of the velocity) as it passes the top of the window is 20.8 m/s.How does the angle of launch affect the kinetic energy of a rubber band?
Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.
State three factors affecting pressure in liquids
Answer:
Density of liquid
Depth of liquid
Acceleration due to gravity
